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Standard XII

Chemistry

Chemical Reactivity

Dioxygen O2...

Question

Dioxygen $(O_{2})$ molecule is stable whereas disulphur $(S_{2})$ is unstable at the room temperature:

due to small size of oxygen, hence strong p-p bonding is possible in

O_{2}

(O=O)

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due to larger size of sulphur, so there is no strong p-p bonding in

S_{2}

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both A & B

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none of these

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Solution

The correct option is A both A & B
In $O_{2} (O = O)$ , due to the small size of oxygen, strong p-p bonding is possible, but in $S_{2}$ , due to the larger size of sulphur, there is no strong p-p bonding. So, bond is weak and breaks down.

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Similar questions

Q. Assertion: Both rhombic and monoclinic sulphur exist as

S_{8}

but oxygen exists as

O_{2} .

Reason: Oxygen form

p π - p π

multiple bond due to small size and small bond length but

p π - p π

bonding is not possible in sulphur due to its larger atomic size than oxygen.

Q. o-hydroxybenzadehyde is a liquid at room temperature while p-hydroxybenzaldehyde is high melting solid because there is intramolecular H-bonding in o-hydroxybenzaldehyde (weak) while intermolecular H-bonding in p-isomer (strong).
if true enter 1. else enter 0.

Q. The formation of a double bond in an oxygen molecule is explained below. Arrange the given points in a sequential order.
(a) 2p

_{y}

orbitals of each oxygen atom overlap laterally/sidewise to form a pi bond.
(b) Thus, a double bond between two oxygen atoms in which one p

_{z}

- p

_{z}

sigma bond and p

_{y}

- p

_{y}

π

bond is formed.
(c) All the three 2p orbitals are perpendicular to each other. Hence 2p

_{z}

orbitals of each oxygen atom overlap end to end to form a sigma bond.
(d) The electronic configuration of oxygen is

O

: 1s

^{2}

^{2}

_{x}^{1}

_{y}^{2}

_{z}^{1}

$S_{1}$ : Bond order of $O_{2}^{-}$ is more than $O_{2}^{+}$
$S_{2}$ : $N O^{+}$ is more stable than $N_{2}^{+}$
$S_{3}$ : $C_{2}$ is more stable than $C_{2}^{+}$
State, in order, whether $S_{1}$ , $S_{2}$ , $S_{3}$ are true or false ?

Q. Fill in the blanks with the appropriate choice.
Bond order of

N_{2}^{+}

P

while that of

N_{2}

O

. Bond order of

O_{2}^{+}

R

while that of

O_{2}

S

N - N

bond distance

T

, when

N_{2}

changes to

N_{2}^{+}

and when

O_{2}

changes to

O_{2}^{+}

, the

O - O

bond distance

U

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Dioxygen (O2) molecule is stable whereas disulphur (S2) is unstable at the room temperature:

The correct option is A both A & BIn O2(O=O), due to the small size of oxygen, strong p-p bonding is possible, but in S2, due to the larger size of sulphur, there is no strong p-p bonding. So, bond is weak and breaks down.

Dioxygen $(O_{2})$ molecule is stable whereas disulphur $(S_{2})$ is unstable at the room temperature:

The correct option is A both A & B
In $O_{2} (O = O)$ , due to the small size of oxygen, strong p-p bonding is possible, but in $S_{2}$ , due to the larger size of sulphur, there is no strong p-p bonding. So, bond is weak and breaks down.